var result = string.Join('，'，ModelState.Values。SelectMany(v => . errors).Select(a=> . errormessage));

var x = new Dictionary<string,string>();
for (var b = 0; b < ViewData.ModelState.Values.Count(); b++)
{
    if (ViewData.ModelState.Values.ElementAt(b).Errors.Count() > 0)
        x.Add(ViewData.ModelState.Keys.ElementAt(b), String.Join(",", ViewData
            .ModelState.Values.ElementAt(b).Errors.Select(c => c.ErrorMessage)));
}

仅仅输出错误消息本身对我来说是不够的，但是这样做达到了目的。

var modelQuery = (from kvp in ModelState
                  let field = kvp.Key
                  let state = kvp.Value
                  where state.Errors.Count > 0
                  let val = state.Value?.AttemptedValue ?? "[NULL]"

                  let errors = string.Join(";", state.Errors.Select(err => err.ErrorMessage))
                  select string.Format("{0}:[{1}] (ERRORS: {2})", field, val, errors));

Trace.WriteLine(string.Join(Environment.NewLine, modelQuery));

使用字段名和错误消息获取错误

var errors = new List<ErrorDto>();
foreach (KeyValuePair<string, ModelStateEntry> kvp in context.ModelState)
{
    if (kvp.Value.Errors.Count > 0)
    {
        errors.Add(new ErrorDto()
        {
            FieldName = kvp.Key,
            ErrorMessage = string.Join(",", kvp.Value.Errors.Select(v => v.ErrorMessage))
        });
    }
}

误差模型

    public class ErrorDto
    {
        public string FieldName { get; set; }
        public string ErrorMessage { get; set; }
    }

这个也有用:

var query = from state in ModelState.Values
    from error in state.Errors
    select error.ErrorMessage;
var errors = query.ToArray(); // ToList() and so on...

我发现，按照到目前为止给出的答案中的建议，可以在不设置错误消息的情况下获得发生异常的情况，因此要捕获所有问题，确实需要同时获取ErrorMessage和Exception。

String messages = String.Join(Environment.NewLine, ModelState.Values.SelectMany(v => v.Errors)
                                                           .Select( v => v.ErrorMessage + " " + v.Exception));

或者作为一种扩展方法

public static IEnumerable<String> GetErrors(this ModelStateDictionary modelState)
{
      return modelState.Values.SelectMany(v => v.Errors)
                              .Select( v => v.ErrorMessage + " " + v.Exception).ToList();

}